Logical problem. . . . I need an immediate answer and ask for help.

Thinking of solving problems 1:

Suppose the number is x, y; The sum is X+Y=A, and the product is x * y = b.

According to what Pang said for the first time, "I'm sure you don't know what these two numbers are." So X+Y is not the sum of two prime numbers. Then the probability of A is 1 1, 17, 23, 27, 29, 35, 37, 4 1, 47, 5 1, 53, 57, 59, 65, 66.

Let's calculate the possible value of b again:

The sum is the product of11:18,24,28,30.

The sum is the product of 17: 30, 42, 52, 60, 66, 70, 72.

The total is the product of 23: 42 and 60. ...

The total is the product of 27: 50 and 72. ...

The sum is the product of 29: ...

The total is the product of 35: 66. ...

The total is the product of 37: 70. ...

......

We can draw the conclusion that the possible B is ... Of course, some numbers (30=5*6=2* 15) appear more than once.

At this time, after comparing and calculating his own figures, Sun said, "I can determine these two figures now."

Based on this sentence and our calculated set of B, we can delete some duplicates from the calculated set of B.

The sum is the product of11:18,24,28.

The sum is the product of 17: 52.

The total is the product of 23: 42 and 76. ...

The sum is the product of 27: 50 and 92. ...

The total is the product of 29: 54 and 78. ...

The total is the product of 35: 96, 124. ...

Sum is the product that 37 can get:, ...

......

Because Pang said, "Since you put it that way, I know what these two numbers are." Then the product obtained by sum must also be unique. From the above, there is only one number left in one line, which is the product of sum 17 52. Then x and y are 4 and 13, respectively.

Problem solving ideas 2:

The voice numbers are S 1, P 1, S2.

Let these two numbers be x, y, and s, product p.

From S 1, P doesn't know these two numbers, so S can't be obtained by adding two prime numbers, S < = 4 1, because if S > 4 1, then P will get 4 1× (S-4 1), so it will definitely be able to.

1). Suppose the sum is 1 1. 11= 2+9 = 3+8 = 4+7 = 5+6. If P gets 18, 18 = 3× 6 = 2× 9, and only 2+9 falls in the set A, then P can be said to be P. Let's see, if P gets 24,24 = 6× 4 = 3× 8 = 2×12, P can also say P 1, because in at least two cases P can say P 1, so A can't assert s 2, so the sum is not1/kloc.

2). Suppose the sum is 17. 17 = 2+ 15 = 3+ 14 = 4+ 13 = 5+ 12 = 6+ 1 1 = 7+ 10 = 8+

3). Suppose the sum is 23. 23 = 2+2 1 = 3+20 = 4+ 19 = 5+ 18 = 6+ 17 = 7+ 16 = 8+ 15 = 9+65438.

4). Suppose the sum is 27. If p gets 8× 19 or 4×23, it can be asserted that P 1, so the sum is not 27.

5). Suppose the sum is 29. If p gets 13× 16 or 7×22, it can be asserted that P 1, so the sum is not 29.

6). Suppose the total number is 35. If p gets 16× 19 or 4×3 1, it can be asserted that P 1, so the sum is not 35.

7). Suppose the sum is 37. If p gets 8×29 or 1 1×26, it can be asserted that P 1, so the sum is not 37.

8). Suppose the sum is 4 1. If b gets 4×37 or 8×33, it can be asserted that P 1, so the sum is not 4 1.

To sum up, these two numbers are 4 and 13.

Ideas to solve the problem 3:

Understanding of Sun Pang's conjecture by hand calculation

1) According to the second half of Pang's first sentence, we are sure that the sum S that Pang knows will definitely not be greater than 54.

Because if and 54

It happens to be 53 and a, so Sun knows that the product m is M=53*a, so Sun knows that there are at least two of these primes.

One contains the factor 53, because 53 is a prime number. However, if it is less than 100 and the factor is 53, it can only be

53 itself, so Sun can infer 53 and a from this product 53*a alone. So if Pang finds out,

If s is greater than 54, he dare not rule out the possibility that two numbers are 53 and A, nor dare he rashly say, "But I'm sure."

You don't know what these two numbers are.

If 53+99

If S=98+99, Pang can immediately judge that these two numbers can only be 98 and 99, and m can only be 98*99.

Sun can also know these two techniques, so this is obviously impossible.

2) According to the second half of Pang's first sentence, we can also determine that what Pang knows and S cannot be expressed as the sum of two prime numbers.

Otherwise, if the two numbers selected by Guiguzi happen to be these two prime numbers, Sun can get a unique prime decomposition after knowing the product m and judge the result. Pang still dared not say, "You must not know what these two numbers are." .

According to Goldbach's conjecture, any even number greater than 4 can be expressed as the sum of two prime numbers. For even numbers below 54, the conjecture must have been verified, so S must not be an even number.

In addition, odd numbers of type S=2+p, where P is an odd prime number, should also be excluded.

And S=5 1 should be excluded, because 51=17+2 *17. If Guiguzi chooses (17,2 *17), then Sun knows.

It will be M=2* 17* 17, and his guess of Guiguzi's original binary number can only be (17,2* 17). (Why 5 1 should be taken out separately depends on the following reasoning. )

3) So we get that S must be in the following numbers:

1 1 17 23 27 29 35 37 4 1 47 53

On the other hand, as long as Pang's S is in these numbers, he can say, "But I'm sure you don't know these two either.

What are numbers? "Because no matter how these numbers are divided into two numbers, at least one number is a composite number (it must be even and even).

Odd number, if even number is greater than 2, it is a composite number. If the even number is equal to 2, our above steps are guaranteed.

Odd numbers are complex numbers), that is, s can only be decomposed into

A) S=2+a*b or b) s = a+2 n * b

In these two ways, a and b are both odd numbers, n >；; = 1。

Then (the two groups of numbers in the "at least two groups of numbers" I said below are different, and they do exist (that is, those)

If the number is less than 100), it will not be written.

A) or the grandson's M=2*a*b, and the grandson will at least be in two groups of numbers (2*a, b) and (2, a*b) (A and.

B is an odd number, so these two groups of numbers must be different);

B) or m = 2 n * a * b,

If n> 1, Sun will at least make up his mind in two groups of numbers (2 (n- 1) * a, 2 * b) and (2 n * a, b);

If n= 1 and a is not equal to b, then the sun is uncertain in at least two groups of numbers (2*a, b) and (2b, a).

Meaning;

If n= 1 and a equals b, which means that S=a+2*a=3a, then s must be a multiple of 3, and we only need to

Only discuss S=27. 27 If it is split into S=9+ 18, then Sun gets M=9* 18, and he is there.

At least (9, 18) and (27, 6) groups can't make up their minds.

(The above discussion of 5 1 is found from the discussion of this last case. I don't know if the above statement is true.

It's too complicated, but looking at the "special case" of 5 1, I suspect that strict argument may be so annoying)

Now we know that if and only if the sum S obtained by Pang is

C={ 1 1， 17，23，27，29，35，37，4 1，47，53}

He would say, "I'm not sure what these two numbers are, but I'm sure you don't know either."

What is this? "

Sun Bin can come to the same conclusion as us. He knows more about that M than we do.

4) Sun's sentence "I can determine these two numbers now" shows that he decomposed M into prime factors and then combined them into.

Among several conjectures about two numbers of Guiguzi, the sum of one and only one conjecture is in C, otherwise he

I still can't make up my mind between many guesses.

After listening to Sun's words, Pang Juan can come to the same conclusion as us. He knows that era better than us.

5) Pang's words "I know what these two numbers are now" indicate that he also got the sum by splitting S into two numbers.

Some conjectures about two numbers of Guiguzi, but among all these disassembly methods, only one meets the requirements in 4).

Conditions, otherwise he won't know what kind of situation, which makes Sun Bin infer these two figures.

So we can exclude those S in C that can be expressed as S = 2 n+p in two ways, where n >；; 1, p is a prime number.

Because if S = 2 N 1+P 1 = 2 N2+P2, whether it is (2 N 1, P 1) or (2 N2, P2), Sun Bin.

The correct result can be determined by m = 2 N 1 * P 1 or m = 2 N 2 * P2, because of various combinations of two numbers obtained by m.

Only with such a combination of (2 n, p) can the sum of two numbers be odd, so it is in C, so Sun Bin can announce that he knows.

What's going on, but Pang Juan still has to worry about (2 N 1, P 1) or (2 N 2, P2).

Because 1 1=4+7=8+3, 23=4+ 19= 16+7, 27 = 4+23 =16+1.

47=4+43= 16+3 1。 So the possible value of s can only be found in

17 29 4 1 53

Medium. Let's continue to shrink this table.

29 is impossible, because 29=2+27=4+25. Whether it is (2, 27) or (4, 25), Sun Bin can correctly judge:

A) If it is (2,27) and m = 2 * 27 = 2 * 3 * 3, then the combination that Sun can guess is (2,27) (3,18) (6,9).

The last two corresponding s are 2 1 and 15, which are not in C, so it is impossible, so it can only be (2,27).

B) If it is (4,25) and M=4*25=2*2*5*5, then the combination that Sun can guess is (2,50) (4,25) (5,20).

( 10, 10)。 Only (4,25) s is in C.

But Pang Juan has to worry about whether Sun Bin's m is 2*27 or 4*25.

4 1 is impossible because 41= 4+37 =10+31. The reasoning behind it is simple.

53 is impossible because 53=6+47= 16+37. The reasoning behind it is simple.

Study 17. Now we have to consider the binary sum and division of all 17:

(2, 15): Then M=2* 15=2*3*5=6*5, 6+5= 1 1 is also in C, so it must not be this m, otherwise 4).

If the conditions are not met, Sun's sentence "I can determine these two figures now" can't be said.

(3, 14): Then m = 3 *14 = 2 * 3 * 7 = 2 * 21,2+2 1=23 is also in C, and the following reasoning is very brief.

(4, 13): Then M=4* 13=2*2* 13. Then the combination that Sun can guess is (2,26) (4, 13), and only (4, 13).

And is in C, so in this case Sun Bin can say the words in 4).

(5, 12): Then M=5* 12=2*2*3*5=3*20, 3+20=23 is also in C, and the following reasoning is very short.

(6, 1 1): then m = 6 *11= 2 * 3 *1= 2 * 33, 2+33=35 is also in C.

(7, 10): Then M=7* 10=2*5*7=2*35, 2+35=37 is also in C, and the reasoning behind it is very brief.

(8,9): Then M=8*9=2*2*2*3*3=3*24, 3+24=27 is also in C, and the reasoning behind it is very brief.

So when S= 17, only in this case (4, 13) can Sun Bin guess what those two numbers are. In this case, Pang Juan knew what these two numbers were and said, "I know what these two numbers are now". After listening to Pang Juan, we also know that these two numbers should be (4, 13).